Library / Foundations: Physics / FPH-66
Foundations: Physics
Equivalent stiffness of springs in parallel
Single spring stiffness that behaves like two springs side by side, both stretched by the same amount.
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- Foundations: Physics
- Result
- keq [N/m]
Springs in parallel stretch by the same amount and their forces add, so their stiffnesses add: k_eq = k₁ + k₂. The combination is always stiffer than either spring. Use the series page when springs are joined end to end.
Single spring stiffness that behaves like two springs side by side, both stretched by the same amount. Their forces add, so the combination is stiffer than either one.
Calculator
- Share of the load carried by spring 10.4
- Natural frequency with a 1 kg mass5.033 Hz
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The two springs together give 1,000 N/m, and spring 1 carries 40 % of any load.
- Stiffer springs take a larger share of the force; each spring stretches by the same amount.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N/m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| k1 Stiffness of spring 1 | 1 to 1,000,000 N/m | 601 | 1,001,000 | 1,000,000 | +10% | +4 % |
| k2 Stiffness of spring 2 | 1 to 1,000,000 N/m | 401 | 1,000,000 | 1,000,000 | +10% | +6 % |
Sweep of Stiffness of spring 1
| Stiffness of spring 1 (N/m) | Equivalent stiffness (N/m) |
|---|---|
| 1 | 601 |
| 5.623 | 605.6 |
| 31.62 | 631.6 |
| 177.8 | 777.8 |
| 1,000 | 1,600 |
| 5,623 | 6,223 |
| 31,620 | 32,220 |
| 177,800 | 178,400 |
| 1,000,000 | 1,001,000 |
Sweep one variable
Variables and how each one contributes
Equivalent stiffness
What the equation gives you. Everything else feeds it.
Stiffness of spring 1
The stiffness of spring 1, in N/m. Raises the equivalent stiffness by its share of the total. Raising it by 10 % (to 440) moves the result by +4 %, from 1,000 to 1,040 N/m.
Stiffness of spring 2
The stiffness of spring 2, in N/m. Raises the equivalent stiffness by its share of the total. Raising it by 10 % (to 660) moves the result by +6 %, from 1,000 to 1,060 N/m.
Worked example with the values above
- Both springs stretch by the same amount, so their forces add and so do their stiffnesses.
- Add the stiffnesses: k_eq = k₁ + k₂ = 400 + 600.
- k_eq = 1,000 N/m.
Assumptions and limits
- Massless springs obeying Hooke’s law.
- Both springs stretched by the same amount, for instance connected to a rigid plate.
Common mistakes
- Using the series formula for springs side by side.
- Forgetting that an offset load can tilt a rigid plate and make the springs stretch unequally.
Statistics and uncertainty
Stiffnesses add, so independent tolerances combine as absolute errors: the relative uncertainty of the sum is smaller than that of each spring when the errors are independent.
Where it is used
- Combining the four springs of a suspension or a set of mounts.
- Finding the natural frequency of a mass on several springs.
Maths you need
- Hooke’s law: F = kx
- Adding forces: springs side by side share the load